Problem 50 - Algebra 2

Question

Find the first five terms of each arithmetic sequence described. \(a_{1}=\frac{5}{8}, d=\frac{3}{8}\)

Step-by-Step Solution

Answer

The first five terms are \(\frac{5}{8}, 1, \frac{11}{8}, \frac{7}{4}, \frac{17}{8}\).

Step 1: Identify the given information

We are given the first term of the arithmetic sequence as \(a_1 = \frac{5}{8}\) and the common difference \(d = \frac{3}{8}\). Our task is to find the first five terms of this sequence.

Step 2: Formula for the nth term

The formula for the nth term of an arithmetic sequence is given by \(a_n = a_1 + (n-1) \cdot d\). This formula will help us to find the subsequent terms of the sequence.

Step 3: Calculate the second term

Using the formula \(a_n = a_1 + (n-1) \cdot d\), find \(a_2\):\[a_2 = a_1 + (2-1) \cdot d = \frac{5}{8} + 1 \cdot \frac{3}{8} = \frac{8}{8} = 1.\] So the second term is 1.

Step 4: Calculate the third term

Now calculate \(a_3\):\[a_3 = a_1 + (3-1) \cdot d = \frac{5}{8} + 2 \cdot \frac{3}{8} = \frac{11}{8}.\] Hence, the third term is \(\frac{11}{8}\).

Step 5: Calculate the fourth term

Calculate the fourth term, \(a_4\):\[a_4 = a_1 + (4-1) \cdot d = \frac{5}{8} + 3 \cdot \frac{3}{8} = \frac{14}{8} = \frac{7}{4}.\] So, the fourth term is \(\frac{7}{4}\).

Step 6: Calculate the fifth term

Finally, calculate \(a_5\):\[a_5 = a_1 + (5-1) \cdot d = \frac{5}{8} + 4 \cdot \frac{3}{8} = \frac{17}{8}.\] Therefore, the fifth term is \(\frac{17}{8}\).

Key Concepts

nth term formula

To unravel the pattern within an arithmetic sequence, the nth term formula is a key tool. This formula is expressed as:
\[ a_n = a_1 + (n-1) \cdot d \]Here,

• \(a_n\) is the term we want to find.
• \(a_1\) denotes the first term of the sequence.
• \(n\) represents the term number.
• \(d\) stands for the common difference between consecutive terms.

This formula enables us to compute any term in the sequence by starting with the first term and adding the common difference multiple times. For example, if you aim to determine the 3rd term, you'd add the common difference twice (because \(3-1 = 2\)) to the first term. This provides a straightforward way to find any specific term in the sequence.

common difference

The common difference is what makes an arithmetic sequence consistent and predictable.In our example, the common difference \(d\) is given as \(\frac{3}{8}\). This tells us how much we need to add to any term to get the next term. It's calculated by subtracting a term from the next term in the sequence.

• For instance, if you know the second term is \(1\) and the first term is \(\frac{5}{8}\), subtracting them yields the common difference.

The arithmetic sequence is said to have a constant step-size set by this difference, which is crucial for maintaining its regular interval. Understanding and identifying this difference allows you to master moving through the sequence confidently.

fraction arithmetic

Working with fractions is pivotal in arithmetic sequences when terms or the common difference aren't whole numbers.Let's break down steps for adding and subtracting fractions, as we see it happening in the exercise:

• Align the denominators: Both terms should share the same bottom number before performing any operation. For instance, when adding \(\frac{5}{8}+\frac{3}{8}\), the denominators are already the same.
• Operate on the numerators: Once the denominators match, you can directly add or subtract the numerators. So, \(5 + 3 = 8\) gives \(\frac{8}{8} = 1\).
• Simplify if possible: Always reduce the fraction to its simplest form, like converting \(\frac{8}{8}\) to \(1\).

Understanding fraction arithmetic greatly enhances your capability to solve arithmetic sequences involving fractions.

sequence terms calculation

Calculating sequence terms systematically is crucial for accurate results.Each term builds on the previous terms using the formula and the common difference.Here’s a recap of how to find each term:

• Start with \(a_1 = \frac{5}{8}\) (the first term).
• Use the nth term formula: \(a_n = a_1 + (n-1) \cdot d\)
• Plug in the term's number you're solving for. For example, for the second term \(a_2\), you use \(n=2\) in the formula to find \(1\).
• Repeat the process for additional terms by adjusting \(n\) accordingly, calculating terms like \(a_3 = \frac{11}{8}\) and \(a_4 = \frac{7}{4}\).

This incremental approach ensures each term flows seamlessly from its predecessor, providing a logical structure to tackle arithmetic sequences efficiently.